Question

The acceleration displacement graph of a particle executing simple harmonic motion is shown in the figure. The time period of simple harmonic motion is

• A. Information is insufficient
  
• B. $4\pi /\sqrt{3}\sec$
• C. The given graph does not represent simple harmonic motion
  
• D. $2\pi /\sqrt{3}\sec$

Answer 1

The correct option is B $4\pi /\sqrt{3}\sec$

As we know,
$a=-{\omega }^{2}x$
As slope of given question is negative, therefore using straight line equation
$y=mx$
For slope calculate the value of $\frac{dy}{dx}$
Here, $y=aandx=x,$ therefore,
$\frac{da}{dx}=\tan {37}^{\circ }$

$\frac{da}{dx}=-\frac{3}{4}$

$a=-\frac{3}{4}x$

As we know,

$a=-{\omega }^{2}x$
${\omega }^2=\frac{3}{2}$

$\frac{2\pi }{T}=\frac{\sqrt{3}}{2}$

$T=\frac{4\pi }{\sqrt{3}}s$

Final answer: (a)